A digit can't be a member of a JExocet if a) it doesn't comply with the 'S' cell requirement and b) it isn't a candidate in either target. If this is the only fly in the ointment then it's an Almost JExocet.

Hi blue, my suggestion is that an Exocet Secondary Equivalence involving a given/solved cell might always lead to the same eliminations as does Exocet Single Target Cell logic. This would only be true for Exocet patterns where SE's are possible, which limits them to 3 row (or column) 3 box patterns. In both of Danny's examples there were in fact 2 SEs, so my suggestion may have to be amended to there being 2 SE's in the pattern, at least one of which involves a given/solved cell. I would think that if 2 logical pathways always lead to the same eliminations there should be a way of proving it, or if they don't always do so, a counter-example could be found.

In writing up the decriptions of JExocets in its various forms I've met this theoretical situation:

The candidates in the base cells are (abcd)The number of cover houses needed to cover the 'S' cells for each digit are (a)=2, (b)=2, (c)=3, (d)=1 This would allow (d) to occupy both target cells if it was paired with (c), and obviously other two digit combinations out of (abc) would also be possible.

Does anyone feel inclined to run a computer search for one of these flying fish?